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1.CONVEX MIRROR-THE MIRROR A SPHERICAL REFLECTING SURFACE (OR ANY REFECTING SURFACE FASIONED INTO A PORTION OF A SPHERE) IN WHICH ITS BULGING SIDE FACES THE SOURCE OF LIGHT.2. CONCAVE MIRROR- Acurved mirror is a mirror with acurved reflecting surface .Explanation:HOPE ITS HELP YOU.....PLEASE MARK IN BRAINLIST ANS WER

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its RESISTANCE will FOURTH of the INITIAL

IT IS USED IN  LAVA LAMPSGENERALLYKNOWN as "FLUID MOTION LAMPS",IT IS AN illustration of density and BUOYANCY in action as the blobs of oozing goo MOVE up and down in ever-changing shapes. These lamps consist of a container of water in which is PLACED a colored organic oily liquid that does not mix with water, thus constituting a second phase. The composition of the oil phase is such that its density is slightly greater than that of water at room temperature, so it normally resides at the bottom of the container. When the lamp is turned on, a heat source  concealed in the base of the container heats the oil phase. This reduces its density to a VALUE below that of the water, causing blobs of oil to rise to the TOP of the container. Being now far removed from the heat source, the blobs cool down and sink back to the bottom, where they repeat the cycle.Explanation:HOPE THIS HELPS YOU

This is the right answerExplanation:I didn't understood the second term and therefore didn't MEAN to confuse you!!!!HOPE THIS HELPS!!!!PLEASE MAKE IT BRAINLIEST ANSWER!!!!

tion: acceleration = 5 cm / sec²initial velocity=u= 0final velocity = V= 25 cm/sectime = t= 5 secondsacceleration = a=?v= u+at25 = 0 + a×5=> a = 25/5 = 5CM /sec²

The burning matchstick would obviously have the higher TEMPERATURE, but the total heat content above ABSOLUTE ZERO would be way higher for the bucket full of hot WATER. Simply stated the very small MASS of the matchstick keeps it from containing very much total heat at all in comparison to the much larger mass of the hot water. Hope that it will help you ✌️

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Calculate coulomb force between two alphaparticles SEPARATED by 3.2×10^-15m?DISTANCE = r =CHARGE on the ALPHA particle = +2e =

Study of fossils is an important aspectof the study of evolution. Fossils are the dead remains or impressions of ORGANISMS which lived in the past. They provide us direct evidence of organic evolution. ... This SHOWS theevolutionary RELATIONSHIP between BIRDS and REPTILES.

No HOLDING a PEN is not a FLUID friction.Mark as BRAINLIEST

ng to the law of CONSERVATION of MECHANICAL energy, whenever there is an intercharge between the POTENTIAL energy and kinetic energy,the total mechanical energy i.e.,K+U=constant when there are no FRICTIONAL forces.

Explanation:OBJECT DISTANCE, U = - 12 CM IMAGE DISTANCE, V = - 48 CM MAGNIFICATION = - (V/U)MAGNIFICATION = - (48/12)MAGNIFICATION = - 4 REAL MAGNIFIED IMAGE IS OBTAINED. HOPE THIS HELPS.

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Derivation of Escape Speed  The derivation starts with the initial gravitational potential energy at the given altitude and the initial kinetic energy of the object. This total initial energy is then compared with the sum of the potential and kinetic energies at an infinite separation, in order to determine the escape velocity equationThe Law of Conservation of Energy states that the total energy of a closed system remains constant. In this CASE, the closed system CONSISTS of the two objects with the gravitational force between them and no outside energy or force affecting either object. Thus the total final energy—potential energy plus kinetic energy—must equal the total initial energy:TEi = TE∞KEi + PEi = 0Substitute values: mve2/2 − GMm/Ri = 0Add GMm/Ri to both sides of equation: mve2/2 = GMm/RiSolve for ve2: ve2 = 2GM/Ri TAKE the square root of each expression to get: ve = ± √(2GM/Ri) Considering our gravitational convention for DIRECTION, ve is upward or away from the other object and is thus negative: ve = − √(2GM/Ri)The derivation of the gravitational escape velocity of an object from a much larger mass is achieved by comparing the potential and kinetic energy values at some given point with the values at infinity, applying the Law of Conservation of Energy. The equation for the gravitational escape velocity is:object from a much larger mass is achieved by comparing the potential and kinetic energy values at some given point with the values at infinity, applying the Law of Conservation of Energy. The equation for the gravitational escape velocity is:ve = − √(2GM/Ri) Taking altitude into account, the equation can be written as: ve = − √[2GM/(R + h)] l ve = − √[2GM/(r + h)]Explanation:

A luminous object is ONE that GIVES off light. In other words, it glows (or SHINES) of its own accord. To be able to GLOW, the object must have its own source of energy.❤

a) VELOCITY is the CORRECT ANSWER.

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The MOTION of OBJECTS is GOVERNED by Newton's LAWS. The same simple laws that govern the motion of objects on earth ALSO extend to the heavens to govern the motion of planets, moons, and other satellites.

The law of conservation of momentum states that for two objects colliding in an ISOLATED system, the total momentum before and after the COLLISION is equal. This is because the momentum LOST by one OBJECT is equal to the momentum gained by the other.plz mark me as BRAINLIEST plz

Hope this help you mate Mark as brainlist The gravitational force between a mass and the Earth is the OBJECT's weight. Mass is CONSIDERED a measure of an object's inertia, and its weight is the force exerted on the object in a gravitational field. On the surface of the Earth, the TWO forces are related by the ACCELERATION due to GRAVITY: Fg = mg.

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no it is not a FLUID FRICTION..

In SI units its VALUE is approximately 6.674×10−11 m3⋅kg−1⋅s−2. The MODERN NOTATION of Newton's law involving G was INTRODUCED in the 1890s by C. V.HOPE IT HELPED U..

tion:EQUIVALENT RESISTANCE will be five .so ACC. to QUESTION .V=IR I=12\5

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Wave frequencyEXPLANATION:The REASON being, this wave EQUATION is of the GENERAL form/boxed{Y=Asin(wt-kx)}Here w=2πf where 'f' is wave frequencyHope this answer helped you ✌️

here is the DERIVATION EXPLANATION:THANK you

In order to ANSWER this question, we must apply newton's first equation of motion v = u + at Where v is the final velocity in m/s, which is we shall attempt to find, u is the initial velocity in m/s, given in the question as 20m/s a is the acceleration of the body in m/s2, in the question, it is indicated to be 4m/s2 t is the time in SECONDS, from the question, it is 4s. v = 20 + 4*4 v = 20 + 16 = 36m/s As for the distance, either the second or third equation of motion can be used. Using the second equation, s (distance) = u*t + 0.5*a*(t^2) s = 20*4 + 0.5*4*(4*4) s = 80 + 32 = 112m Using the third equation of motion, v2 = U2 + 2as 36^2 = 20^2 + 2*4*s 1296 = 400 + 8 * s 8*s = 1296 - 400 = 896 s = 896/8 = 112mExplanation:

re mainly three equations of motion which describe the relationship between velocity, time, acceleration and displacement.First, consider a body moving in a straight line with uniform acceleration. Then, let the initial velocity be u, acceleration be a, time period be t, velocity be v, and the distance travelled be S.The equation of motions derivation can be done in three ways which are:Derivation of equations of motion by Simple Algebraic MethodDerivation of Motion by Graphical MethodDerivation of Motion by CALCULUS MethodBelow, the equations of motion are derived by all the three methods in a simple and easy to understand way.Derivation of First Equation of MotionThe first equation of motion is:v = u + atDerivation of First Equation of Motion by Algebraic MethodIt is KNOWN that the acceleration (a) of the body is defined as the rate of change of velocity.So, the acceleration can be written as:a = v − utFrom this, rearranging the terms, the first equation of motion is obtained, which is:v = u + atDerivation of First Equation of Motion by Graphical MethodConsider the diagram of the velocity-time graph of a body below:Derivation Of Equation Of MotionIn this, the body is moving with an initial velocity of u at point A. The velocity of the body then changes from A to B in time t at a uniform rate. In the above diagram, BC is the final velocity i.e. v after the body travels from A to B at a uniform acceleration of a. In the graph, OC is the time t. Then, a perpendicular is drawn from B to OC, a parallel line is drawn from A to D, and another perpendicular is drawn from B to OE (represented by dotted lines).Following details are obtained from the graph above:The initial velocity of the body, u = OAThe final velocity of the body, v = BCFrom the graph,BC = BD + DCSo, v = BD + DCv = BD + OA (since DC = OA)Finally, v = BD + u (since OA = u) (Equation 1)Now, since the slope of a velocity-time graph is equal to acceleration a,So,a = slope of line ABa = BD/ADSince AD = AC = t, the above equation becomes:BD = at (Equation 2)Now, combining Equation 1 & 2, the following is obtained:v = at + uDerivation of First Equation of Motion by Calculus MethodIt is known that,Derivation Of Equation Of MotionSo,Derivation Of Equation Of MotionDerivation of Second Equation of MotionThe second equation of motion is:S = ut + ½ a2Derivation of Second Equation of Motion by Algebraic MethodConsider the same notations for the derivation of the second equation of motion by simple algebraic method.Derivation Of Second Equation Of MotionDerivation of Second Equation of Motion by Graphical MethodTaking the same diagram used in first law derivation:Derivation Of Equation Of MotionIn this diagram, the distance travelled (S) = Area of figure OABC = Area of rectangle OADC + Area of triangle ABD.Now, the area of the rectangle OADC = OA × OC = utAnd, Area of triangle ABD = (1/2) × Area of rectangle AEBD = (1/2) at2 (Since, AD = t and BD = at)Thus, the total distance covered will be:S = ut + (1/2) at2Derivation of Second Equation of Motion by Calculus MethodVelocity is the rate of change of displacement.Mathematically, this is expressed as\(v=\frac{ds}{dt}\)Rearranging the equation, we get\(ds=vdt\)Substituting the first equation of motion in the above equation, we get\(ds=(u+at)dt\) \(=(udt+at\,dt)\) \(\int_{0}^{s}ds=\int_{0}^{t}u\,dt+\int_{0}^{t}at\,dt\) \(s=ut+\frac{1}{2}at^2\)Derivation of Third Equation of MotionThe third equation of motion is:v2 = u2 + 2aSDerivation of Third Equation of Motion by Algebraic MethodDerivation Of Third Equation Of MotionDerivation of Third Equation of Motion by Graphical MethodDerivation Of Equation Of MotionThe total distance travelled, S = Area of trapezium OABC.So, S= 1/2(SumofParallelSides)×HeightS=(OA+CB)×OCSince, OA = u, CB = v, and OC = tThe above equation becomesS= 1/2(u+v)×tNow, since t = (v – u)/ aThe above equation can be written as:S= 1/2(u+v)×(v-u)/aRearranging the equation, we getS= 1/2(v+u)×(v-u)/aS = (v2-u2)/2aThird equation of motion is obtained by solving the above equation:v2 = u2+2aSDerivation of Third Equation of Motion by Calculus MethodIt is known that,Derivation Of Equation Of MotionThese were the detailed derivations for equations of motion in the graphical method, algebraic method and calculus method.Equations of Motion FormulaEquations of motion FormulaFirst equation of motion v=u+atSecond equation of motion \(s=ut+\frac{1}{2}at^{2}\)Third equation of motion v2 = u2+2as

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he LENGTH of the lour hand is 14 cm.To find:Displacement of the hour hand in 9 hours.Solution:1) The length of the hour hand would be the radius of the circular PATH covered by the hand of the clock.2) Hand covers the TIME of 9 hours so it will COVER 3/4 of the total distance.3) So the displacement will be 3/4(2πr)3/2×πr3/2×22/7×1466 cm.Displacement of the hour hand in 9 hours is 66 cm.

Gauss's law states that the net flux of an electric FIELD in a closed SURFACE is directly PROPORTIONAL to the enclosed electric CHARGE. ... It was initially formulated by Carl Friedrich Gauss in the year 1835 and relates the electric fields at the POINTS on a closed surface and the net charge enclosed by that surface.

tion:given that the NET potential ENERGY of the SYSTEM is EQUAL to ZERO

"High VOLTAGE and TWO METAL plates /ELECTRODES Semolina and OIL in container Connect a (high) voltage to the plates in container Semolina lines up in the fieldSource:

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